The percentage composition by weight of an aqueous solution of a solute (molecule mass 150) which boils at 373.26 is

mol .  mass = 150

${\mathrm{T}}_{\mathrm{b}}=373.26$

${\mathrm{K}}_{\mathrm{b}}=0.52$

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\times \mathrm{m}$

$\frac{0.26}{0.52}=m$

$\mathrm{m}=\frac{1}{2}$

$\frac{1}{2}\times 150=75g$ Wt of the solute

no. of moles of ${\mathrm{H}}_2\mathrm{O}=0.5$

Weight of solvent = 1000g .

1075g  of solvent contain 75g of the solute

100g of solvent contain − ?

$\frac{100\times 75}{1075}=7\%$